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The correct expression for partial pressure in terms of mole fraction is:
A. \[{{\text{p}}_1} = {{\text{x}}_1}{{\text{P}}_{{\text{total}}}},{{\text{p}}_2} = {{\text{x}}_2}{{\text{P}}_{{\text{total}}}}\]
B. \[{\text{p}} = {{\text{x}}_1}{{\text{x}}_2}{{\text{P}}_{{\text{total}}}}\]
C. \[{{\text{P}}_{{\text{total}}}} = {{\text{p}}_1}{{\text{x}}_1},{{\text{P}}_{{\text{total}}}} = {{\text{p}}_2}{{\text{x}}_2}\]
D. \[{{\text{p}}_1} + {{\text{p}}_2} = {{\text{x}}_1} + {{\text{x}}_2}\]

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Last updated date: 13th Jun 2024
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Answer
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Hint: Dalton’s law of partial pressure is the principle behind this question. It states that total pressure of a mixture of non-reacting gases in a system is equal to the sum of their partial pressure. Ideal gas equation is also applied here and thus the relation between partial pressure and mole fraction is calculated.

Complete answer:
Partial pressure is the pressure exerted if it were present alone and occupied the same volume as in the mixture. Dalton’s law of partial pressure explains about the total pressure of a mixture of gas in a container is related with the partial pressure of each gases.
\[{{\text{P}}_{{\text{total}}}} = {{\text{p}}_1} + {{\text{p}}_2} + {{\text{p}}_3} + ....\]
\[{{\text{P}}_{{\text{total}}}}\]is the total pressure of the mixture of gases.
\[{{\text{p}}_1},{{\text{p}}_2}{\text{,}}{{\text{p}}_3},....\]are the partial pressures of each gases.
Consider a container having two gases-1 and 2. The gas has a volume \[{\text{V}}\], temperature \[{\text{T}}\], number of moles \[{\text{n}}\], gas constant \[{\text{R}}\]. These gases exert partial pressure \[{{\text{p}}_1},{{\text{p}}_2}\] on the wall of the container.
Using ideal gas equation, \[{\text{PV}} = {\text{nRT}}\]

For gas 1, \[{{\text{p}}_1} = \dfrac{{{{\text{n}}_1}{\text{RT}}}}{{\text{V}}} \to \left( 1 \right)\]
\[{{\text{n}}_1}\] is the number of moles of gas 1.
For gas 2, \[{{\text{p}}_2} = \dfrac{{{{\text{n}}_2}{\text{RT}}}}{{\text{V}}} \to \left( 2 \right)\]
\[{{\text{n}}_2}\] is the number of moles of gas 2.
\[{{\text{P}}_{{\text{total}}}} = {{\text{n}}_{{\text{total}}}}\dfrac{{{\text{RT}}}}{{\text{V}}} \to \left( 3 \right)\], where \[{{\text{n}}_{{\text{total}}}} = {{\text{n}}_1} + {{\text{n}}_2}\]
Dividing equation \[\left( 1 \right)\]by \[\left( 3 \right)\], we get
\[\dfrac{{{{\text{p}}_1}}}{{{{\text{P}}_{{\text{total}}}}}} = \dfrac{{{{\text{n}}_1}{\text{RT}}}}{{\text{V}}} \div \dfrac{{{{\text{n}}_{{\text{total}}}}{\text{RT}}}}{{\text{V}}}\]
Cancelling the common terms, we get
\[\dfrac{{{{\text{p}}_1}}}{{{{\text{P}}_{{\text{total}}}}}} = \dfrac{{{{\text{n}}_1}}}{{{{\text{n}}_{{\text{total}}}}}} \to \left( 4 \right)\]
Mole fraction can be defined as the ratio of number of moles of each gas to the total number of moles of mixture of gases.
\[\therefore {{\text{x}}_1} = \dfrac{{{{\text{n}}_1}}}{{{{\text{n}}_{{\text{total}}}}}}\], where mole fraction is represented by ${{\text{x}}_1}$.
Now equation $\left( 4 \right)$can be written as,
${{\text{p}}_1} = {{\text{x}}_1}{{\text{P}}_{{\text{total}}}}$
Therefore the option A is correct.

Additional Information:
Dalton’s law of partial pressure has many applications. When a water-insoluble gas is collected over water, the gas becomes moist. Total pressure of moisture will be equal to the sum of pressure of gas and water.

Note:
Partial pressure of oxygen in outside air is higher than partial pressure of air in lungs. Thus, at higher altitudes, we feel difficulty in breathing since the pressure is low. Pilots may have uncomfortable breathing in a non-pressurized cabin.